Respuesta :
Answer:
Explanation:
Consider the diagram for the charges on the given sphere(check attachment).
The electric field at this point are
E(r) = 0 for r≤a. Equation 1
E(r) = kq/r² for a<r<b. Equation 2
E(r) = 0 for b<r<c. Equation 3
E(r) = kq/r² for r>c. Equation 4.
We know that electric potential is related to the electric field using
V = Ed
A. The potential at outer surface of the hollow sphere (at r=c) can be calculated as,
The electric field at this point is
E = kQ / r²
Then,
V = Ed,
At d = r = c
Then,
Vc = (kQ / c²) × c
Vc = kQ / c
Then, Q has charges +q, -q and +q
Then, Q = q - q + q = q
V = kq / c
B. The potential at inner surface of the hollow sphere (at r=b) can be calculated as,
V = kQ/r
V = kQ / b, since r = b
Then, Q = q
V = kq / b
C. At r = a
Then, from equation 1.
E(r) = 0 for r≤a. Equation 1
The electric field at the surface of the solid sphere is 0, E = 0N/C
Then,
V = Ed = 0 V
So the electric potential at the surface of the solid sphere is 0
D. At r = 0
Then, electric potential can be calculated using
V = kq / r
Then, r = 0
V = kq / 0
V → ∞

In this exercise we have to use the knowledge of a spherical shell, like this:
A)[tex]Vc = kQ / c[/tex]
B)[tex]V = kq / b[/tex]
C) [tex]0 V[/tex]
D)[tex]V \rightarrow \infty[/tex]
Recalling some important equations to perform this exercise we have:
- [tex]E(r) = 0 \ for \ r\leq a[/tex]
- [tex]E(r) = kq/r^2 \ for \ a
- [tex]E(r) = 0 \ for \ b
- [tex]E(r) = kq/r^2 \ for \ r>c[/tex]
A) Calculating the shell potential we have:
[tex]E = kQ / r^2\\V = Ed\\d = r = c\\Vc = (kQ / c^2) * c\\Vc = kQ / c[/tex]
B) The potential at inner surface of the hollow sphere:
[tex]V = kQ/r\\V = kq / b[/tex]
C) Calculating the value of the potential when we have the radius in infinity, like this:
[tex]E(r) = 0 \\V = Ed = 0 V[/tex]
D) Calculating the value of the potential when we have the radius at zero, like this:
[tex]V = kq / r\\V = kq / 0\\V \rightarrow \infty[/tex]
See more about potencial at brainly.com/question/2701710
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